+ 8

I felt difficulty in solving coding problems on pre and post increment problems

difference in pre/ post increment operator problem solving technique.

4th Jul 2017, 6:51 PM
KUSHAGRA GUPTA
KUSHAGRA GUPTA - avatar
10 Respostas
+ 8
thanks to all for nice explanations
1st Aug 2017, 2:56 AM
KUSHAGRA GUPTA
KUSHAGRA GUPTA - avatar
+ 7
@JoJo - thanks for the clarification. I kinda get what your saying, and I understand incrementation a bit more now because of it, but I think I'll have to experiment with it in some practice codes to really grasp it fully.
8th Aug 2017, 5:49 AM
Boo Da Man
Boo Da Man - avatar
+ 4
I also had some difficulty with this. I get the concept, but where specifically does "i" get incremented when, let's say, a = i++ People have said in their answers for this question, after the next "statement", or after the next "sentence"; is this referring to the next line of code? And I think someone said, the next time "i" is used. Someone please explain specifically where the variable gets incremented when using post-incrementation.
5th Aug 2017, 6:04 PM
Boo Da Man
Boo Da Man - avatar
+ 3
thanks @jojo and Boo DaMan
8th Aug 2017, 11:26 AM
KUSHAGRA GUPTA
KUSHAGRA GUPTA - avatar
+ 2
Let i=2 ++i means increment before execution of sentence. so a=++i results a=3. Now again i=2 i++ indicates increment after execution of sentence. a=i++ will make a=2 and will increment i. so i becomes 3
4th Jul 2017, 7:00 PM
Shashank Shinde
Shashank Shinde - avatar
+ 2
small suggestion: avoid loosing to much time on things like "what's the result of ++a+++b++.....". things like that are never used and considered bad practice as their solutions are compiler dependant.
5th Jul 2017, 4:39 AM
seamiki
seamiki - avatar
+ 2
When you write : int x=0, y; y=(--x)+(--x)+(--x); C++ thinks this : y=(--x)+(--x); y=y+(--x); Soooo : y=(--x)+(--x) ;// -2+-2=-4 y=y+(--x);// -4+-3=-7 --x //-1 --x //-2 x+x // x=-2 --x //-3 -2+x // x=-3 inctementation is fantastic but it becomes quick very very complicated as you can see. If you don't understand ask me, don't hesitate!
5th Aug 2017, 9:26 PM
Jojo
+ 1
++i increment i's value before anything : int a=5, i=4; cout << a==++i; // Same as: i+=1; cout << a==i; i++ increment i's value after ONE statement, the one where i is used: int a=5, i=4; cout << a==i++; // same as: cout << a==i; i+=1; If you have other questions on incrementation ask !
4th Jul 2017, 7:03 PM
Jojo
+ 1
I think what they forget to mention is that the first variable then changes. I found it hard to get my head around i=3. if you imagine i=2 a = ++i to me this means a is 3 but i is still 2. But No, the ++ increments i aswell. So both i and a are equal to 3
24th Jul 2017, 5:05 PM
emily
+ 1
You're right the term 'statement' must be defined. In my answer, a 'statement' is an 'action' accomplished by the compiler : a=1; is one statement. a=a+1; is composed of two statements in C++. Let say a=1. C++ read this : a+1 ---> temp // I used '--->' to show that C++ remembers the value of 'a+1' So this was the first statement. then : a=temp //remember that 'temp' is not a real variable, I used it to show how it works. This is the second statement. In other words : a=a+1;//C++ evaluate first a+1. Then, it evaluates a=... (here it is the value of 'a+1') Ask if you don't understand
5th Aug 2017, 6:38 PM
Jojo